The statement is that $C_g$ and $C_h$ are "equal, up to a movement". In his proof the author replaces $C_h$ by a congruent copy (again denoted by $C_h$) in the following way: He chooses a $p\in E$ and applies a rotation $R$ of ${\mathbb R}^3$ such that the original orthonormal triple $\bigl(T_h(p),N_h(p),B_h(p)\bigr)$ is mapped to the triple $\bigl(T_g(p),N_g(p),B_g(p)\bigr)$. When this constant rotation $R$ is applied to $C_h$ the curve $R(C_h)=:C_h$ does not yet coincide with $C_g$, but is (in fact) a translate of $C_g$. When you want you can apply in addition a translation $A$ such that $(A\circ R)(h(p))=g(p)$ , but it is not necessary. As readers we accept without further ado that the moved curve $C_h$ is congruent to the original $C_h$.
The hard part of the proof then consists in showing that the new $C_h$ is congruent to $C_g$. Here the Frenet formulas are used. You should actually compute $\phi'$ in order to see that the equality of $s\mapsto\kappa(s)$ and $s\mapsto\tau(s)$ for the two curves plays a role in showing that $\phi'=0$:
$$\eqalign{\phi'&=(T_g\cdot T_h+N_g\cdot N_h+B_g\cdot B_h)'\cr
&=T_g'\cdot T_h+T_g\cdot T_h'+N_g'\cdot N_h+N_g\cdot N_h'+B_g'\cdot B_h+B_g\cdot B_h')\cr
&=\kappa N_g\cdot T_h+\kappa T_g\cdot N_h+(-\kappa T_g+\tau B_g)\cdot N_h+(-\kappa T_h+\tau B_h)\cdot N_g-\tau N_g\cdot B_h-\tau B_g\cdot N_h\cr &=0\ .\cr}$$
In the end the "equality" of $C_g$ and $C_h$ comes from the uniqueness part for the solution of ODEs.
I followed the comments and got the following answer:
We have that $\alpha$ is parametrized by arch lenght with $\kappa (s_0)=0$ and $\kappa'(s_0) \neq 0$.
Observations:
If $\kappa(s_0) \neq 0$ then the curve do not intersect $\alpha$'s tangent line.
If $\kappa(s_0) = 0$ we have that $\alpha$ not necessarily cross the tangent line. For example, take $\alpha(t) = (t^3, t^6)$ at $s_0=0$, we have $\kappa (s_0)=0$ and the curve cross the tangent line.
Thus, the conditions for the curve to cross the tangent line - case $\kappa(s_0) = 0$ -, is that $\kappa'(s_0) \neq 0$.
Supose $\alpha$ parametrized by arc lenght. If we want that curve cross the line
$$\alpha(s) - \alpha(s_0) = \alpha(s) T(s_0) + B(s)N(s_0)$$
were $T=\alpha'$, $B$ is the binormal and $N$ the normal vector of $\alpha$, we need to study the sign of $B(S)$ were
$$B(s) = \langle \alpha(s)-\alpha(s_0), N(s_0) \rangle.$$
By Taylor we have
$$\alpha(s) = \alpha(s_0) + (s-s_0) \alpha'(s_0) + \frac{(s-s_0)^2}{2} \alpha''(s_0) + \frac{(s-s_0)^3}{3!} \alpha'''(s_0) + R(s_0).$$
How $\|\kappa (s_0) \|=\|\alpha''(s_0)\|=0$ then
$$\alpha(s) - \alpha(s_0) = \alpha'(s_0)(s-s_0) + \frac{(s-s_0)^3}{3!} \alpha'''(s_0) + R(s_0)$$
with
$$\lim_{s \to s_0} \frac{R(s_0)}{(s-s_0)^3}=0.$$
Replacing in $B(s) = \langle \alpha(s)-\alpha(s_0), N(s_0) \rangle$ we have
$$B(s) = (s-s_0) \underbrace{\langle \alpha'(s_0), N(s_0)\rangle}_{=0} + \frac{(s-s_0)^3}{3!} \langle \alpha'''(s_0), N(s_0)\rangle + \langle R(s_0), N(s_0)\rangle$$
and
$$B(s) = \frac{(s-s_0)^3}{3!} \langle \alpha'''(s_0), N(s_0)\rangle + \langle R(s_0), N(s_0)\rangle.$$
By Frenet formula $T(s) = \kappa(s) N(s) \Rightarrow T'(s) = \kappa'(s)N(s) + \kappa(s) N'(s)$ then
$$T'(s_0) = \kappa'(s_0) N(s_0)$$
and
$$B(s) = \frac{(s-s_0)^3}{3!} \langle \kappa'(s_0) N(s_0), N(s_0)\rangle + \langle R(s_0), N(s_0)\rangle.$$
Let
$$\frac{B(s)}{(s-s_0)^3} = \frac{1}{3!} \langle \kappa'(s_0) N(s_0), N(s_0)\rangle + \left \langle \frac{R(s_0)}{(s-s_0)^3}, N(s_0)\right\rangle \xrightarrow[s\to s_0]{} \underbrace{\frac{1}{3!}}_{>0} \cdot \kappa'(s_0) \underbrace{\|N(s_0)\|}_{>0}$$
then the signal of $B(s)$ depends only of $\kappa'(s_0)$.
Case 1: If $\kappa'(s_0) >0$ then $B(s)>0$ and so there is a neighborhood of $s_0$ such that $\langle \alpha(s) - \alpha(s_0), N(s_0)\rangle > 0$.
Case 2: If $\kappa'(s_0) <0$ then $B(s)<0$ and so there is a neighborhood of $s_0$ such that $\langle \alpha(s) - \alpha(s_0), N(s_0)\rangle < 0$.
Therefore, for all neighborhood of $s_0$ there are points in both half planes determined by the tangent line to $\alpha$ at $s_0$.
Best Answer
Your parametrizations $f$ and $g$ are not parametrizations by path-length, and so the theorem you quote does not apply to them. Moreover, even if they were, the hypothesis $k_f(s)=k_g(s)\neq 0$ for all $s$ does not hold, since you your calculation the curvatures are $0$ at $t=0$.